Suppose I have $n$ column vectors of equal length $\{\vec{a}_i\}_{i \in \{1,...,n\}}$ which are linearly independent of one another. Suppose I have a further $n$ column vectors $\{\vec{b}_i\}_{i \in \{1,...,n\}}$, each of the same length as $\{\vec{a}_i\}_{i \in \{1,...,n\}}$, which I know are also linearly independent of one another. Finally, suppose I now know that:
$$\sum_i \vec{a}_i'\vec{b}_i = 0$$
Is this information enough to conclude any of the following results? If so why/why not?
- $\vec{a}_i=\vec{0}$ for all $i$.
- $\vec{b}_i=\vec{0}$ for all $i$.
- $\vec{a}_i'\vec{b}_i=\vec{0}$ for all $i$.
(Where $\vec{0}$ is the zero vector of appropriate dimension).
I'll stick to two dimensions for ease of notation, then you can consider the following: $$ a_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad a_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \qquad b_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad b_2 = \begin{bmatrix} 0 \\ -1 \end{bmatrix} $$
Then $a_1^Tb_1+a_2^Tb_2 = 1-1 = 0$ and yet none of $a_i$, $b_i$, nor $a_i^Tb_i$ are zero, despite them being linearly independent pairs of vectors.